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A force of 100 N acts at a point P(-2, 3, 5)m has its line of action through Q(10, 3, 4)m. Calculate moment of this force about origin (0, 0, 0)
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written 8.4 years ago by
teamques10
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• modified 8.4 years ago |
$$S^2= (∆x^2 + ∆y^2 + ∆z^2) $$
$ S^2 = ((-2-10)^2+(3-3)^2 + (5-4)^2) \\ s = 12.05\\ Fx = F∆x/s = 100 \times (-12)/12.05 = -99.6 \\ Fy = F∆y/s = 100 \times (0)/12.05 = 0 \\ Fz = F∆z/s = 100 \times (1)/12.05 = 8.3 $
So, in vector form
$F = -99.6i + 8.3 k \\ . r= vector \space \space P – vector \space \space O = (-2, 3, 5) = -2i +3j +5k \\ M = r \times F \\ = \text {determinant} $
$= -24.9i + 481.4j - 298.8k Nm$
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